Question at position 11 find ∫23−xdx\int2^{3-x}dx−e3−xln⁡(2)+C- \frac{e^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C \frac{2^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C\frac{2^{3 - x}}{\ln(2)} + C−23−xln⁡(2)+C- \frac{2^{3 - x}}{\ln(2)} + C

Question

BlackTom AI · Accepted Answer

First, I’ll restate the problem pieces to ensure we’re analyzing the same expressions: the integral in question is ∫ 2^{3 − x} dx, and we’re evaluating multiple displayed forms of the antiderivative.

Option A: -(e^{3 − x}) / ln(2) + C. This form uses e^{3 − x} rather than 2^{3 − x}. Since the base is e rather than 2, this is not the correct antiderivative for 2^{3 − x}. The derivative of e^{3 − x} involves e^{3 − x} itself, not 2^{3 − x}, so the constants and base don’t align with the original integrand. In short, confusing the base from 2 to e leads to an incorrect result.

Option B: 2^{3 − x} / ln(2) + C. This resembles the standard integral ∫ a^{u} du = a^{u}/ln(a) when u is the variable of integration and a is constant. However, we must account for the inner function u = 3 − x, whose derivative du/dx = −1. This introduces a negative sign, so the correct antiderivative should carry a minus sign to compensate for du = −dx. Therefore this form misses the necessary negative sign and is not correct.

Option C: 2^{3 − x} / ln(2) + C. This is identical in mathematical content to Option B (same expression with the same missing negative sign issue). As explained, ignoring the minus from the chain rule when substituting u = 3 − x yields an incorrect result.

Option D: −2^{3 − x} / ln(2) + C. This form includes the required negative sign that comes from the chain rule when integrating with respect to x after setting u = 3 − x. Concretely, ∫ 2^{3 − x} dx = ∫ 2^{u} (−du) = − ∫ 2^{u} du = − 2^{u} / ln(2) + C = − 2^{3 − x} / ln(2) + C. Thus the negative sign correctly accounts for du = −dx, and the 2^{3 − x} reflects the original base and exponent structure.

Putting these together, the only option that properly incorporates both the base 2 and the minus sign from the substitution is the one that yields −2^{3 − x} / ln(2) + C. The other options either replace the base with e, or omit the essential negative sign required by the chain rule, leading to an incorrect antiderivative.

Question at position 11 find ∫23−xdx\int2^{3-x}dx−e3−xln⁡(2)+C- \frac{e^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C \frac{2^{3 - x}}{\ln(2)} + C 23−xln⁡(2)+C\frac{2^{3 - x}}{\ln(2)} + C−23−xln⁡(2)+C- \frac{2^{3 - x}}{\ln(2)} + C单项选择题

类似问题

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Question at position 7 ∫(x3−1x4+2)dx=\int\left(x^3-\frac{1}{x^4}+2\right)dx=x44−3x3+2x+C\frac{x^4}{4}-\frac{3}{x^3}+2x+Cx44+13x3+2x+C\frac{x^4}{4}+\frac{1}{3x^3}+2x+C3x2+4x−5+C3x^2+4x^{-5}+C3x2−14x3+C3x^2-\frac{1}{4x^3}+Cx44−13x3+2x+C\frac{x^4}{4}-\frac{1}{3x^3}+2x+C

Question at position 4 ∫2x−13dx=\int\frac{2x-1}{3}dx=(2x−1)26+C\frac{\left(2x-1\right)^2}{6}+C13(x2−x)+C\frac{1}{3}\left(x^2-x\right)+C12+C\frac{1}{2}+Cx2−x3x+C\frac{x^2-x}{3x}+C23+C\frac{2}{3}+C

Question at position 7 ∫(x3−1x4+2)dx=\int\left(x^3-\frac{1}{x^4}+2\right)dx=x44−13x3+2x+C\frac{x^4}{4}-\frac{1}{3x^3}+2x+Cx44+13x3+2x+C\frac{x^4}{4}+\frac{1}{3x^3}+2x+Cx44−3x3+2x+C\frac{x^4}{4}-\frac{3}{x^3}+2x+C3x2+4x−5+C3x^2+4x^{-5}+C3x2−14x3+C3x^2-\frac{1}{4x^3}+C

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